Suppose the quadratic function $f\left( x \right) =a{ x }^{ 2 }+bx+c$ is such that $f\left( -2 \right) =0\quad and\quad \frac { -b }{ 2a } =1$, solve $f\left( x \right)=0$.

The equation f(-2) = 0 means x = -2 is one root of the quadratic. The expression -b/2a = 1 represents the axis of symmetry, meaning the midpoint of the roots is 1. If one root is -2, the other root must be positioned symmetrically across 1, calculated as 2*1 - (-2) = 4. Thus, the solutions to f(x) = 0 are x = -2 and x = 4.

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